Week 7 Hypothesis Testing Q & A
1. Steve listens to his favorite streaming music service when he works out. He wonders whether the
service algorithm does a good job of finding random songs that he will like more often
...
Week 7 Hypothesis Testing Q & A
1. Steve listens to his favorite streaming music service when he works out. He wonders whether the
service algorithm does a good job of finding random songs that he will like more often than not. To
test this, he listens to 50 songs chosen by the service at random and finds that he likes 32 of them.
Use Excel to test whether Steve will like a randomly selected song more than not and then draw a conclusion in the
context of a problem. Use α = 0.05. Typeequationhere .
Ho: p = ≤ 0.5 (50%) p = 0.5
Ha: p = > 0.5 (strictly ¿≠ )
P-value = 0.02 which is < α=0.05 we reject Ho and support the Ha
Hypothesis Test for p population
proportion
Level of Significance 0.05
(decimal
)
Proportion under H0 0.5000
(decimal
)
n 50
Number of
Successes 32
Sample Proportion
0.64000
0
StDev
0.50000
0
SE
0.07071
1
Test Statistic (z)
1.97989
9
One-Sided p-value
0.02385
2
Two-Sided p-value
0.04770
4
Right-Tailed (>) 1.644854
Left-Tailed (<) -1.644854
Two-Tailed (≠) ± 1.959964
Answer: Reject the null hypothesis. There is sufficient evidence to prove that Steve will like a random selected song
more often than not.
2. A magazine regularly tested products and gave the reviews to its customers. In one of its reviews, it
tested 2 types of batteries and claimed that the batteries from company A outperformed batteries
from company B in 108 of the tests. There were 200 tests. Company B decided to sue the magazine,
claiming that the results were not significantly different from 50% and that the magazine was
slandering its good name.Use Excel to test whether the true proportion of times that Company A’s batteries outperformed Company B’s batteries is
different from 0.5. Identify the p=value rounding it to 3 decimal places.
Ho: p = 0.5 Ha ≠ 0.5 (two tailed test) n = 200 (α is not given soleave it 0.05)
Hypothesis Test for p population
proportion
Level of Significance 0.05
Proportion under H0 0.5000
n 200
Number of Successes 108
Sample Proportion 0.540000
StDev 0.500000
SE 0.035355
Test Statistic (z) 1.131371
One-Sided p-value 0.129238
Two-Sided p-value 0.258476
Right-Tailed (>) 1.644854
Left-Tailed (<)
-
1.644854
Two-Tailed (≠) ± 1.959964
Answer: 0.258 (because it is a two tailed test). We are not rejecting the null hypothesis and we do not have evidence to
support the alternative hypothesis.
3. A candidate in an election lost by 5.8% of the vote. The candidate sued the state and said that more
than 5.8% of the ballots were defective and not counted by the voting machine, so a full recount
would need to be done. His opponent wanted to ask for the case to be dismissed, so she had a
government official from the state randomly select 500 ballots and count how many were defective.
The official found 21 defective ballots.
Use Excel to test if the candidates claim is true and that < 5.8% of the ballots were defective. Identify the p=value
rounding to 3 decimal places.
Ho: p = ≥ 0.058 Ha ¿0.058 (one tailed test) n = 500 (α is not given soleave it 0.05)
Hypothesis Test for p population
proportion
Level of Significance 0.05 (decimal)
Proportion under H0 0.0580 (decimal)
n 500
Number of Successes 21
Sample Proportion 0.042000
StDev 0.233743
SE 0.010453
Test Statistic (z) -1.530613
One-Sided p-value 0.063008
Two-Sided p-value 0.126016
Right-Tailed (>) 1.644854
Left-Tailed (<) -1.644854
Two-Tailed (≠) ± 1.959964
Answer: 0.063
4. A researcher claims that the incidence of a certain type of cancer is < 5%. To test this claim, a
random sample of 4000 people are checked and 170 are found to have the cancer.
The following is the set up for the hypothesis:
Ho = 0.05
Ha = < 0.05
In the example the p-value was determined to be 0.015.
Come to a conclusion and interpret the results of this hypothesis test for a proportion (use a significance level of 5%)
Answer: The decision is to reject the null hypothesis. The conclusion is that there is enough evidence to support the
claim.
5. A researcher is investigating a government claim that the unemployment rate is < 5%. TO test this
claim, a random sample of 1500 people is taken and it is determined that 61 people were
unemployed.
Ho: p = 0.05 Ha: p < 0.05
Find the p-value for this hypothesis test for a proportion & round to 3 decimal places.
Hypothesis Test for p
population proportion
Level of Significance 0.05
Proportion under H0 0.0500
n 1500
Number of
Successes 61
Sample Proportion
0.04066
7
StDev
0.21794
5
SE
0.00562
7
Test Statistic (z) -
1.658577
One-Sided p-value
0.04845
7
Two-Sided p-value
0.09691
4
Answer: 0.048
6. An economist claims that the proportion of people that plan to purchase a fully electric vehicle as
their next car is greater than 65%.
To test this claim, a random sample of 750 people were asked if they planned to purchase a fully electric vehicle as their
next car. Of this 750, 513 indicated that they plan to purchase an electric vehicle.
Ho: p = 0.65 Ha; p = >0.65
Find the p-value for this hypothesis test for a proportion & round to 3 decimal places.
Hypothesis Test for p
population proportion
Level of Significance 0.05
Proportion under H0 0.6500
n 750
Number of
Successes 513
Sample Proportion
0.68400
0
StDev
0.47697
0
SE
0.01741
6
Test Statistic (z)
1.95217
5
One-Sided p-value
0.02558
8
Two-Sided p-value
0.05117
6
Answer: 0.026
7. Colton makes the claim to his classmates that < 50% of newborn babies born this year in his state
are boys. To prove this claim, he selects a random sample of 344 birth records in his state from this
year. Colton found that 176 of the newborns were boys. What are the null and alternative
hypothesis for this hypothesis test.
Answer: Ho: 0.5Ha: <0.5
8. An Airline company claims that in its recent advertisement that at least 94% of passenger luggage
that is lost is recovered and reunited with their customer within 1 day. Hunter is a graduate student
studying statistics. For a research project, Hunter wants to find out whether there is sufficient
evidence in support of the airline company’s claim. He randomly selects 315 passengers whose
luggage was lost by the airlines and found out that 276 of those passengers were reunited with their
luggage within 1 day. Are all of the conditions for his hypotheses test met, and if so, what are the Ho
and Ha for this hypothesis test?
For a binomial Model to follow the normal model, the following condition must be satisfied:
Success count = n * p ≥ 5 and Failure count ≥5
Example: success count 315 * 0.94 = 296.1 and failure count 315-296.1 = 18.9 so it meets the conditions.
Answer: All of the conditions were met and the Ho = 0.94; Ha = >0.94
9. A college administrator claims that the proportion of students who are nursing majors is > 40%. To
test this claim, a group of 400 students are randomly selected and its determined that 190 are
nursing majors. The following is the set up for the hypothesis test: Ho: p = .40 and Ha: p = >.40
Find the test statistics for this hypothesis test for a proportion & round to 2 decimal places.
Answer: 3.06
Level of
Significance 0.05
Proportion under
H0 0.4000
n 400
Number of
Successes 190
Sample Proportion 0.475000
StDev 0.489898
SE 0.024495
Test Statistic (z) 3.061862
One-Sided p-value 0.001107
Two-Sided p-value 0.002214
10. A hospital administrator claims that the proportion of knee surgeries that are successful are 87%. To
test this claim, a random sample of 450 patients who underwent knee surgery is taken and it is
determined that 371 patients had a successful knee surgery operation. Ho: p = 0.87 Ha: p ≠ 0.87
(two sided tail)
Find the test statistics for this hypothesis test for a proportion & round to 2 decimal places.
Answer: -2.87 (this would be rejected)
Level of 0.05Significance
Proportion under
H0 0.8700
n 450
Number of
Successes 371
Sample Proportion 0.824444
StDev 0.336303
SE 0.015853
Test Statistic (z) -2.873534
One-Sided p-value 0.002052
Two-Sided p-value 0.004104
11. Jose, a competitor in cup stacking, has a sample stacking time mean of 7.5 seconds from 13 trials.
Jose still claims that his average stacking time is 8.5 seconds, and the low average can be contributed
to chance. At the 2% significant level, does the data provide sufficient evidence to conclude that
Jose’s mean stacking time is less than 8.5 seconds? Given the sample data below, select or reject the
hypothesis. (If p=value is < alpha value, we would automatically reject the hypothesis)
Ho: μ = 8.5 Ha: μ = <8.5
α = 0.02 (significance level)
Zo = -2.18
P = 0.0146
Answer: Reject the null hypothesis because the p value 0.0146 is less than the significance level 0.02
12. Marty, a typist, claims his average typing speed is 72 wpm. During a practice session, Marty has a
sample typing speed mean of 84 wpm based on 12 trials. At the 5% significance level, does the data
provide sufficient evidence to conclude that his mean typing speed is >72 wpm? Accept or reject the
hypothesis given the data below.
Ho: μ=72wpm ; Ha: μ=¿72wpm ; α=0.05 (significance level) ; Zo = 2.1; p = 0.018
Answer: Reject the null hypothesis because the p-value 0.018 is less than the significance level α=0.05
13. What is the p-value of a right-tailed one mean hypothesis test, with a test statistic of Zo = 2.1? (Do
not round your answer. Compute your answer using a value from the table. (Value in table was
0.982)
1 – 0.982 = p=value of 0.018
Answer: 0.018
14. What is the p-value of a two-tailed one mean hypothesis test, with a test statistic of Zo = 0.27? (Do
not round your answer. Compute your answer using a value from the table. (Value in table was
0.606)1-0.606 = P-value of 0.394 because it is two tailed, you multiply 2 x .394 = .788
Answer: 0.788
15. Raymond, a typist, claims his average typing speed is 89 wmp. During a practice session, Raymond
has a sample typing speed mean of 95.5 wmp based on 15 trials. At the 1% significance level, does
the data provide sufficient evidence to conclude that his mean typing speed is > 89 wmp? Accept or
reject the hypothesis given the sample data below:
Ho: μ=89 ; Ha: μ=¿89 wmp α=0.02 Sig Level; Zo = 2.75; p = 0.003
Answer: Reject the null hypothesis because p-value 0.003 is less than alpha 0.02
16. Kurtis is a statistician who claims that the average salary of an employee in the city of Yarmouth is
no more than $55,000 per year. Gina, his colleague, believes this to be incorrect, so she randomly
selects 61 employees who work in Yarmouth and record their annual salary. Gina calculates the
sample mean income to be $56.500 per year with a sample standard deviation of $3750. Using the
alternative hypothesis, Ha = μ=¿ $55,000 , find the test statistic τ and the p-value for the
appropriate hypothesis test. Round the τ to 2 decimal places and the p-value to 3 decimal places.
Hypothesis Test for µ for unknown
population stdev
Level of
Significance 0.1
(decimal
)
Mean under H0 55000
n 61
Sample Mean 56500
StDev 3750
SE
480.13830
0
Test statistic (t) 3.124100
One-Sided p-value 0.001373
Two-Sided p-value 0.002746
Answer: t = 3.12; p-value = 0.001
17.A college administrator claims that the proportion of students that are nursing majors is
less than 40%. To test this claim, a group of 400 students are randomly selected and its
determined that 149 are nursing majors.
The following is the setup for this hypothesis test:
H0:p=0.40
Ha:p<0.40
Find the p-value for this hypothesis test and round your answer to 3 decimal places.Hypothesis Test for p
population proportion
Level of Significance 0.05
Proportion under H0 0.4000
n 400
Number of
Successes 149
Sample Proportion
0.37250
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